Question Number 196000 by Rodier97 last updated on 15/Aug/23
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{lim}_{{x}\rightarrow+\infty} \:\left({lnx}\right)^{\mathrm{2}} −\sqrt{{x}}\:\:\:\:\:???? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by MM42 last updated on 15/Aug/23
$${tip}\:\::\:\:{lim}_{{x}\rightarrow+\infty} \:\frac{{lnx}}{{x}^{{p}} }=\mathrm{0}\:\:;\:{p}>\mathrm{0} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\left({ln}^{\mathrm{2}} {x}−\sqrt{{x}}\right)={lim}_{{x}\rightarrow+\infty} \:\sqrt{{x}}\left(\frac{{lnx}}{\:\sqrt[{\mathrm{4}}]{{x}}}\:−\mathrm{1}\right)\left(\frac{{lnx}}{\:\sqrt[{\mathrm{4}}]{{x}}}+\mathrm{1}\right) \\ $$$$=+\infty×−\mathrm{1}×\mathrm{1}=−\infty \\ $$$$ \\ $$
Answered by Frix last updated on 15/Aug/23
$${x}=\mathrm{e}^{\mathrm{2}{t}} \:\mathrm{leads}\:\mathrm{to}\:\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\:\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{e}^{{t}} \right)\:\mathrm{and}\:\mathrm{obviously} \\ $$$$\mathrm{e}^{{t}} \:\mathrm{grows}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{4}{t}^{\mathrm{2}} \:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\infty \\ $$
Commented by MM42 last updated on 15/Aug/23
$${ok} \\ $$